Math Is Fun Forum

coolwind

Member

Registered: 2005-10-30

Posts: 30

A sinusoidal problem~~

If a signal X(t)=5cos(2pi60t)-3sin(2pi180t)

let Cn=(1/T)∫X(t)e^(-jwnt) dt   where T=1/60 , w=2(pi)f   f=1/T

find the Cn ?

Thank you~~

polylog

Member

Registered: 2006-09-28

Posts: 162

Re: A sinusoidal problem~~

Well I think it looks like you just have to carry out the integral.

But first:

w=2(pi)f = 2pi * 60 = 120pi

Cn= (1/T) ∫ X(t)e^(-jwnt) dt

  = (1/T) ∫ [5cos(2pi60t)-3sin(2pi180t)]*e^(-jwnt) dt

  = 60 ∫ [ 5cos(120pi t)e^(-j(120pi)nt) - 3sin(360pi t)e^(-j(120pi)nt) ] dt

These 2 integrals are fairly simple using integration by parts, but its just messy numbers.

polylog

Member

Registered: 2006-09-28

Posts: 162

Re: A sinusoidal problem~~

Actually you can make this a little more pleasant by using the formulas:

cos(t) = ( e^(jt) + e^(-jt) )/2

sin(t) = ( e^(jt) - e^(-jt) )/(2j)

Thus:

5cos(120pi t) = 5/2 e^(j120pi t) + 5/2 e^(-j120pi t)

3sin(360pi t) = -(3/2)j e^(j360pi t) + (3/2)j e^(-j360pi t)

Now we can write x(t) as:

x(t) = 5/2 e^(j120pi t) + 5/2 e^(-j120pi t) +
       (3/2)j e^(j360pi t) - (3/2)j e^(-j360pi t)

so now Cn is as below (I'll use "INT" for the integral sign):

Cn= (1/T) INT x(t)e^(-jwnt) dt
  = INT 60 x(t)e^(-j120pi nt) dt

now the integrand will be the product of e^(-j120pi t) with the 4 terms in the expression for x(t) above.

And the integration will be very easy.

The integrand is:

60 x(t)e^(-j120pi t) =
150 e^( j120pi t - j120pi nt ) +
150 e^( -j120pi t - j120pi nt ) +
90j e^( j360pi t -j120pi nt ) -
90j e^( -j360pi t -j120pi nt )

=

150 e^( j120pi(1-n)t ) +
150 e^( -j120pi(1+n)t ) +
90j e^( j120pi(3-n)t ) -
90j e^( -j120pi(3+n)t ) 

Now this is easy to integrate, so Cn is:

Cn = -j150/(120pi(1-n)) e^( j120pi(1-n)t ) +

     j150/(120pi(1+n)) e^( -j120pi(1+n)t ) +

     90/(120pi(3-n)) e^( j120pi(3-n)t ) -

     90/(120pi(3+n)) e^( -j120pi(3+n)t ) .

But that's still rather ugly... if anyone has a nicer solution, please post.